When 12.00 moles of potassium chlorate decomposes, how many dm3 of oxygen are produced at 325K and 188 kPa? 2KClO3 →2KCl + 3O2
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258.71 dm³

Explanation:

We'll begin by calculating the number of mole of O₂ produced by the decomposition of 12 moles of KClO₃. This can be obtained as follow:

The balanced equation for the reaction is given below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 12 moles of KClO₃ will decompose to produce = (12 × 3)/2 = 18 moles of O₂.

Finally, we shall determine the volume of the O₂. This can be obtained as follow:

Temperature (T) = 325 K

Pressure (P) = 188 KPa

Number of mole (n) = 18 moles

Gas constant (R) = 8.314 KPa.dm³/Kmol

Volume (V) =?

PV = nRT

188 × V = 18 × 8.314 × 325

188 × V = 48636.9

Divide both side by 188

V = 48636.9 / 188

V = 258.71 dm³

Thus, 258.71 dm³ of oxygen were obtained from the reaction.

it means there are 3 atoms of oxygen